MCQ Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 9

What mass of

$NaCl$ would contain the same total number of ions as 245 g of

$MgCl_2$ ?

0%
245 g
0%
225 g
0%
263 g
0%
None of these

Explanation

$(24+35.5\times2)$ gm of

$mgcl_{2}\rightarrow 3$ No ion

245 gm of

$mgcl_{2}\rightarrow \dfrac{3}{(24+70)}\times 245\,gm$

$(\dfrac{3}{24+70}) \times 245$ ion of Nacl

$\rightarrow \dfrac{(23+35.5)}{2\times Na} \times \dfrac{3}{(24+70)} \times 245\,gm$

1176478_1035774_ans_b67cc40c0e244f998f8ebab4fd9dd25e.jpeg

$AgNO_3$ sample is 85% by mass. To prepare

$12.5\ m$ of 0.05 molar

$AgNO_3$ solution,

$AgNO_3$ sample required is:

0%
2.36 g
0%
1.25 g
0%
1.49 g
0%
none of these

Explanation

Given, percentage purity of

$\mathrm{AgNO}_{3}$ Sample

$=85 \%$

We are to prepare

$125 \mathrm{~mL}$ solution haring molarity

$=0.05$ .

Thus, moles of pure

$\mathrm{AgNO}_{3}$ needed = Volume

$\times$ molarity

$\begin{array}{l}=\dfrac{125}{1000} \times 0.05\mathrm{mo}\\=6.25\times 10^{-3} \mathrm{~mol} .\end{array}$

weight of pure

$\operatorname{AgN} O_{3}$ needed

$=\left(6.25 \times 10^{-3}\right) \times 170\mathrm{gm}$

$=1.0625 \mathrm{gm} .$

Let, mass of

$AgNO_3$ sample required =

$m$ gm.

$\begin{array}{l}\text { Thus, } m \times \dfrac{85}{100}=1.0625 \\\therefore \quad m=1.0625\times\dfrac{100}{85}\mathrm{gm}=1.25\mathrm{gm}\end{array}$

So, correct option is (B)

A fresh

$H_2O_2$ solution is labeled as

$11.2V$ . Calculate its concentration in wt/vol percent.

0%
$3.03$
0%
$6.8$
0%
$1.7$
0%
$13.6$

Explanation

$Density = \dfrac{v}{11.2} * M.wt$

$d = \dfrac{11.2}{11.2} * 34 = 34$

$\dfrac{wt}{v} = \dfrac{34}{11.2} = 3.03%$

According to the

$CaCO_3+HCl\rightarrow CaCl_2 +CO_2 + H_2O$ . What mass of

$CaCO_3$ is required to react completely with

$25$ ml. of

$0.75M$

$HCl$ ?

0%
0.9945 g
0%
0.6645 g
0%
0.9375 g
0%
0.8555 g

Explanation

Using morality, Number of moles are calculated for

$1000\;ml$ or

$1\;L$ of the solution.

$1000\;ml$ contains

$0.75\;M \;HCl$

$HCl=0.75$ mole.

Therefore,

$25\;ml$ of

$0.75\;M$ HCl will contains

$HCl =\dfrac{0.75 \times 25}{1000}=0.01875\;mole$

$2$ moles of

$HCl$ reacts with

$=1\;mole$ of

$CaCO_3$

Therefore,

$0.01875$ mole of

$HCl$ will react with

$=\dfrac{1 \times 0.01875}{2}$

$=0.009375\;mole$

Molar mass of

$CaCO_3=100\;g$

Hence, the mass of

$0.009375$ of

$CaCO_3=no.\;of\;moles \times molar\;mass=0.9375\;g$

Two flasks

$A$ and

$B$ of

$500$ mL each are respectively filled with

$O_2$ and

$SO_2$ at

$300$ K and

$1$ atm pressure. The flasks contain:

0%
the same number of atoms
0%
the same number of molecules
0%
more number of moles in flask $A$ as compared to flask $B$
0%
the same mass of gases

Explanation

According to Avogadro's Hypothesis: Equal volumes of all gases contain equal number of atoms under similar conditions of temperature and pressure. So, the correct option will be '

$A$ '.

15 g of methyl alcohol is present in 100 mL of solution. If the density of solution is 0.96 g

${ mL }^{ -1 }$ , Calculate the mass percentage of methyl alcohol in solution.

0%
45.23%
0%
15.66%
0%
25.36%
0%
None of these

Explanation

Mass of methyl alcohol = 15 g

Volume of the solution = 100 ml

Density =

$0.96 gm L^-$

$^1$

So. Density =

$\dfrac{mass}{volume}$

mass = 100

$\times$ 0.90 = 96 g

Mass percentage of methyl alcohol =

$\dfrac{ mass\ of\ methyl\ alcohol}{mass\ of\ solution}$

$\times$ 100

$\dfrac{1500}{96}=15.66\%$

A gas cylinder was found unattended in a public place. The investigating team took the collected samples from it. The density of the gas was found to be

$2.380\ g L^{-1}$ at

$15^oC$ and pressure. Hence the molar mass of the gas is:

0%
30
0%
71
0%
32
0%
58

Explanation

Applying ideal gas law,

$PV = nRT$

$P = \dfrac{wt}{Mwt\times Volume}\times RT$

$Mwt = \dfrac{d}{P}\times RT = \dfrac{2.380}{1}\times 0.0821\times 288 = 58$

What is the mass of the precipitate formed when

$50$ mL of

$16.9$ % solution of

$AgNO_3$ is mixed with

$50$ mL of

$5.8$ %

$NaCl$ solution ?
(Ag =

$107.8$ , N =

$14$ , O =

$16$ , Na =

$23$ ,

$Cl = 35.5$ )

0%
$7$ g
0%
$14$ g
0%
$28$ g
0%
$3.5$ g

Explanation

$50mL\,\, 16.9\%\,AgNO_3=\dfrac{50\times 16.9}{100\times 169}=0.5mol\,AgNO_3$

$50mL\,\,15.8\%\,NaCl=\dfrac{50\times 5.8}{100\times 5.8}=0.05mol\, NaCl.$

$AgNO_3\underset{aq}{+}NaCl\underset{aq}{\rightarrow} AgCl\downarrow +NaNO_3(aq)$

$0.05$

$0.05$ - -

- -

$0.05$

molecular mass of

$AgCl$ =143.3

mass of ppt

$(AgCl)$ =

$0.05\times 143.3$

$=7.1gm$

Hence, option

$A$ is correct.

Pehal and Ishaan were making

$100g$ sugar solution with concentration by mass

$20\%$ and

$50\%$ respectively. Pehal added

$20 g$ of sugar in her solution whereas ishaan evaporated

$20 g$ of water from his solution. Now, the solutions were mixed to form a final solution, The concentration by mass of final solution is:

0%
$28\%$
0%
$45\%$
0%
$40\%$
0%
$60\%$

Explanation

For Pehal:- a mass of sugar=

$20g$

mass of water=

$80g$

The new mass of sugar=

$20+20=40g$

$\therefore$ Concentration of sugar=

$\cfrac { 40 }{ 120 } \times 100=33.33$ %

For Ishaan:- mass of sugar= mass of water=

$50g$

New mass of water=

$50-20=30g$

$\therefore$ Concentration of sugar=

$\cfrac { 50 }{ 80 } \times 100=62.5$ %

Concentration by mass of final solution=

$\cfrac { New\ mass\ of\ sugar\ in\ Pehal+Mass\ of\ sugar\ in\ Ishaan }{ New\ mass\ of\ solution\ in\ Pehal+New\ mass\ of\ solution\ in\ Ishaan }$

$\cfrac { 40+50 }{ 120+80}$

$45$ %

3.49g of ammonia at STP occupies a volume of 4.48

${ dm }^{ 3 }$ calculate the molar mass of ammonia.

0%
13.25 g/mol
0%
14.59 $g/mol$
0%
17.45 $g/mol$
0%
None of these

Explanation

Volume = 4.48

$dm^3$ = 4.48 L

Gram Molecular Mass = 22.4 L at STP

So,

4.48 L of ammonia gas weighs = 3.49 g

22.4 L of ammonia weighs =

$\frac{22.4 \times 3.49}{4.48}$

= 17. 45 g / mol

Two isotopes of an elements

$Q$ are

${ Q }^{ 97 }$ (23.4% abundance) and

${ Q }^{ 94 }$ (76.6% abundance).

${ Q }^{ 97 }$ is

$8.082$ times heavier than

${ C }^{ 12 }$ and

${ Q }^{ 94 }$ is

$7.833$ times heavier than is

${ C }^{ 12 }$ . What is the average atomic weight of the elements

$Q$ ?

0%
$94.702$
0%
$78.913$
0%
$96.298$
0%
$94.695$

Explanation

Average mass

$=\dfrac{\text{Abundance of} \ Q^{97}\times \text{mass of}\ Q^{97} \times \text{Abundance of}\ Q^{94} \times \text{mass of}\ Q^{94}}{100}$

$=\dfrac{23.4\times 8.082\times 12+76.6\times 7.833\times 12}{100}$

$=94.702$ Answer.

$1$ g-atom of nitrogen atom represents:

0%
$6.02\times { 10 }^{ 23 } \ N_2$ molecules
0%
$22.4\ L$ of ${ N }_{ 2 }$ at $S.T.P$
0%
$11.2\ L$ of $N_2$ at $S.T.P.$
0%
$28 g$ of nitrogen

Explanation

$1g-atom$ of nitrogen atom = 1 mole of nitrogen atom

$1$ mole of nitrogen atom have

$6.02\times 10^{23}$ atoms and occupies

$11.2\ L$ of

$N_2$ at STP.

A solution of ethanol in water is 10% by volume. If the solution and pure ethanol have densities of 0.9866 g/cc and 0.785 g/cc respectively. The percent by weight is nearly?

0%
7.95%
0%
17%
0%
9.86%
0%
16.2%

Explanation

Volume of ethanol = 10 ml

Volume of solution = 100 ml

Weight of ethanol =

$Volume \times density$

$10 \times 0.785$

= 7.85 g

Weight of solution =

$100 \times 0.9866$

= 98.66 g

Weight percent of ethanol =

$\frac{7.85}{98.66}$

$\times 100$

$7.95\%$

$10g$ of glucose is dissolved in

$150g$ of water. The mass percantage of glucose is:

0%
$2.50\%$
0%
$6.67\%$
0%
$8.75\%$
0%
$10\%$

Explanation

$10g$ is dissolved in

$150g$ of water

$\therefore$ Mass percentage of glucose=

$\cfrac {10}{150}\times 100$

$=6.67$ %

$N_{2}$ gas is present in one litre flask at a pressure of

$7.6 \times 10^{-10}\ mm$ of

$Hg$ . The number of

$N_{2}$ gas molecules in the flask at

$O^{o}C$ is?

0%
$2.68 \times 10^{9}$
0%
$2.68 \times 10^{10}$
0%
$1.34 \times 10^{28}$
0%
$2.68 \times 10^{22}$

Explanation

To calculate the number of moles of the gas under the given condition by the relation

$PV=nRT$

$P=7.6\times { 10 }^{ -10 }$

$=\dfrac { 7.6\times { 10 }^{ -10 } }{ 760atm } =1\times { 10 }^{ -12 }atm$

$V=1\quad litre$

$T=273+0=273K$

$R=0.082\quad litr\quad atm/K/mol$

substituting the values in equation

$n=\dfrac { PV }{ RT } =\dfrac { 1\times { 10 }^{ -12 }\times 1 }{ 0.082\times 273 } moles$

Now since 1mole=

$6.023\times { 10 }^{ 23 }$

$\dfrac { { 10 }^{ -12 } }{ 0.082\times 273 } =\dfrac { 6.023\times { 10 }^{ 23 }\times { 10 }^{ -12 } }{ 0.082\times 273 }$

$=2.7\times { 10 }^{ 10 }molecule$

If one gram of

$S$ contains

$x$ atoms, atoms is one gram

$O$ will be:

0%
$x$
0%
$x/2$
0%
$\dfrac {2}{3}x$
0%
$2x$

Explanation

$1$ gm of

$S=\cfrac {1}{32}$ moles

$=x$ atoms

$1$ gm of

$O=\cfrac {1}{16}$ moles

$\therefore$ Atoms in

$1$ gm of

$O=\cfrac {x\times \cfrac {1}{16}}{\cfrac {1}{32}}=2x$

In a test tube, there is

$18\ g$ of glucose

$(C_{6}H_{12}O_{6})0.08$ mole of glucose is taken out. Glucose left in the test tube is:

0%
$0.10\ g$
0%
$0.02\ g$
0%
$0.10\ mole$
0%
$3.60\ g$

Explanation

Moles of glucose=

$\cfrac {18}{180}=0.1$ moles

Moles of glucose left=

$0.1-0.08=0.02$ moles

$\therefore$ Mass of glucose left=

$0.02\times 180$

$3.6g$

Rearrange the following in the order of increasing mass and choose the correct order (Atomic mass,

$N=14,O=16,Cu=63$ )
(I) 1 molecule of oxygen
(2) 1 atom of nitrogen
(3)

$1\times {10}^{-10}gm$ molecule of oxygen
(4)

$1\times {10}^{-10}gm$ of copper

0%
$II< I< IV< III$
0%
$IV< III< II< I$
0%
$II< III< I< IV$
0%
$III< IV< I< II$

Explanation

(1)

$1$ molecule of oxygen=

$O_2$

$\therefore$ Mass of

$O_2=\cfrac {16\times 2}{N_A}=\cfrac {32g}{N_A}=\cfrac {32}{6.022\times 10^{23}}=5.3\times 10^{-23}g$

(2)

$1$ atom of Nitrogen=

$\cfrac {1}{N_A}$ moles

$=\cfrac {1}{6.022\times 10^{23}}$ moles

$={1.66\times 10^{-24}}$ moles

$\therefore$ Mass of

$1$ atom of Nitrogen=

$1.66\times 10^{-24}\times 14$

$=23.2\times 10^{-24}g$

(3)

$1\times 10^{-10}gm$ molecule of oxygen=

$1\times 10^{-10}$ moles of

$O_2$

$\therefore$ Mass of

$1\times 10^{-10}gm$ molecule of oxygen=

$1\times 10^{-10}\times 32$

$=3.2\times 10^{-9}g$

(4) Mass of copper

$=1\times 10^{-10}gm$

Comparing the masses in (1), (2), (3) and (4)

we get,

$(2)<(1)<(4)<(3)$

Therefore, answer is

$(II)<(I)<(IV)<(III)$ .

The number of

$g$ molecule of oxygen in

$6.0\times {10}^{24}$ molecules of

$CO$ gas is:

0%
$1g$ molecule
0%
$0.5g$ molecule
0%
$5g$ molecule
0%
$10g$ molecule

Explanation

Number of g molecule of

$O=$ Number of mole of

$O$ gas

Number of moles of

$CO$

$=\dfrac { 6.022\times { 10 }^{ 24 } }{ 6.022\times { 10 }^{ 23 } } =10$

No. of moles of oxygen atom

$=10$

No. of g molecule of

$O$ is

$\dfrac { 10 }{ 2 } =5$ g

$1.5 gm$ mixture of

${ SiO }_{ 2 }$ and

${ Fe }_{ 2 }{ O }_{ 3 }$ on very strong heating leave a residue weighting

$1.46 gm$ . The reaction responsible for loss of weight is

${ Fe }_{ 2 }{ O }_{ 3 }(s)\longrightarrow { Fe }_{ 3 }{ O }_{ 4 }(s)+{ O }_{ 2 }(g)$
What is the percentage by mass of

${ Fe }_{ 2 }{ O }_{ 3 }$ in original sample?

0%
$80\%$
0%
$20\%$
0%
$40\%$
0%
$60\%$

Explanation

$3 Fe_2 O_3 \rightarrow 2 Fe_3 O_4 + \dfrac{1}{2} O_2 (g)$

Molecular mass of

$Fe_2 O_3 = 160$

Molecular mass of

$Fe_3 O_4 = 232$

Loss in mass per

$480 gm \, Fe_2 O_3 = 16 gm$

In this ques, loss of mass =

$0.04 gm$

Mass of

$Fe_2 O_3 = \dfrac{0.04}{16} \times 480$

$= 1.20 gm$

$\therefore \%$ mass of

$Fe_2 O_3 = \dfrac{1.20}{1.50} \times 100 = 80\%$

Calculate number of electrons present in

$9.5\ g$ of

$PO_{4}^{3-}$ .

0%
$6\ N_{A}$
0%
$4.7\ N_{A}$
0%
$5\ N_{A}$
0%
$0.1\ N_{A}$

Explanation

$Mass\ of\ PO_4^{3-}=9.5$

$g$

$Number\ of\ moles =\cfrac{Weight \ of \ PO_4^{3-}}{Molecular\ weight} =\cfrac{9.5}{95}=0.1$

$1\quad mole=6.023\times 10^{23} ions$

$\Rightarrow 0.1\quad moles=0.1\times 6.023\times 10^{23}=6.023\times 10^{22} ions$

Each ion of

$PO_4^{3-}$ has

$50$ electrons

$\therefore$ Total number of electrons

$= 6.023\times 10^{22}\times 50$

$=3\times 10^{24}$ electrons

$= 6.023\times 10^{22}\times \cfrac{100}{2}$

$=6.023\times 10^{23}\times \cfrac{10}{2}$

$=(N_A)5$

$=5N_A$

The density of a

$3 \,M$ sodium thiosulphate

$(Na_2S_2O_3)$ solution is

$1.25 \,gm/ml$ . Calculate the percentage by weight of sodium thiosulphate.

0%
$73.92$
0%
$65.84$
0%
$17.14$
0%
$37.92$

Explanation

Percentage by weight of sodium thiosulphate solution

$(Na_2S_2O_3)$ will be:

$=\cfrac{wt.\ of\ Na_2S_2O_3}{wt.\ of\ solution}\times 100$

Also wt. of

$4Na_2SO_3=$ Molarity

$\times$ molecular wt.

$=3\times 150$

$\%$ by wt.

$=\cfrac{3\times 158}{1.25\times 100}\times 100=37.92$

The mass of

$50\%(w/w)$ solution of

$HCl$ required to react with

$100\ g$ of

$CaCO_{3}$ would be?

0%
$73\ g$
0%
$100\ g$
0%
$146\ g$
0%
$200\ g$

Explanation

The reaction is:

$2HCl+CaCO_3\longrightarrow CaCl_2+CO+3H_2$

$2$ molecules of

$HCl$ give

$1$ molecule of

$CaCO_3$

(Molar mass)

$_{CaCO_3}=100$

$g$

$100$

$g$ of

$CaCO_3=1$

$mole$

So, Molar mass of

$HCl=35.5+1=36.5$

$\therefore 1$

$mole$ of

$HCl=36.5$

$g$

Since, we get

$2$

$moles$ of

$HCl$ from

$100$

$g(1$

$mole)$ of

$CaCO_3$

Weight of

$HCl=2(36.5)$

$g=73$

$g$

Equivalent wt.of a metal is

$2.5$ times higher than oxygen. The ratio of weight of metal to the oxide is:

0%
$0.4$
0%
$1.4$
0%
$2.4$
0%
$3.4$

Explanation

Solution:- (A) 0.4

Given that the equivalent weight of a metal is

$2.5$ times higher than oxygen.

Equivalent weight of oxygen

$= 8$

$\therefore$ Equivalent weight of metal

$=$ equivalent wt. of oxygen

$+ \; (2.5 \; \times$ equivalent weight of oxygen)

$= 8 + 2.5 \times 8 = 28$

As we know that

$\text{eq. wt.} = \cfrac{\text{mol. wt. } \left( {M}_{m} \right)}{\text{valence factor } \left( n \right)}$

$\therefore$ Molecular weight of metal

$\left( {M}_{m} \right) = 28 n$

Formula of metal oxide will be-

${M}_{2}{O}_{n}$

Molecular weight of metal oxide

$= 2 \times {M}_{m} + n \times {M}_{O} = 56n + 16 n = 72 n$

Ratio of weight of metal to its oxide

$= \cfrac{\text{wt. of metal}}{\text{wt. of metal oxide}} = \cfrac{28n}{72n} = 0.389 \approx 0.4$

Hence the required answer is

$(A) 0.4$ .

"One gram moecule of gas at N.T.P occupies

$22.4$ litres." This fact was derived from?

0%
Dalton's therory
0%
Avogadro's hypothesis
0%
Berzelius hypothesis
0%
Law of gaseous volume

One mole of a mixture of CO and

${ CO }_{ 2 }$ requires exactly 20 g of NaOH in solution for complete conversion of all the

${ CO }_{ 2 }$ into

${ Na }_{ 2 }{ CO }_{ 3 }$ . How much NaOH would it require for conversion into

${ Na }_{ 2 }{ CO }_{ 3 }$ , if the mixture (one mole) is completely oxidized to

${ CO }_{ 2 }$ ?

0%
60 g
0%
80 g
0%
40 g
0%
20 g

Explanation

let

$x$ and

$y$ mass of

$CO$ and

$CO_{2}$ be there

$x+y=1$

$2NaOH+CO^{2}\rightarrow Na_{2}CO_{3}+H_{2}O$

$n$ of

$NaOH=20/40=0.5$

$\therefore 0.5/2$ mol

$CO_{2}$ must be there

$\Rightarrow 0.25$ mol

$CO_{2}$

$\therefore 0.75$ mol

$CO$

$CO$ is oxidised to

$CO_{2}$

$NaOH$ required

$=2\times 0.75\times 40$

$=60g$

Vapour density of the equilibrium mixture of

${ NO }_{ 2 }$ and

${ N }_{ 2 }{ O }_{ 4 }$ is found to be 40 for the equilibrium:

${ N }_{ 2 }{ O }_{ 4 }\rightleftharpoons 2{ NO }_{ 2 }$ .

Calculate the percentage of

${ NO }_{ 2 }$ in the mixture?

0%
10%
0%
5%
0%
26.08%
0%
None of these

Explanation

$N_2O_4\rightleftharpoons 2NO_2$

$t=0$

$1$

$t=t_{eq}\ 1-\alpha \quad 2\alpha$

Initial vapour density of N

$_2$ O

$_4$

$=\cfrac{92}{2}=46$

Vapour density at equilibrium=40

$\cfrac{46}{40}=\cfrac{c+c\alpha}{c}$

$\cfrac{46}{40}=1+\alpha$

$\dfrac {46}{1+\alpha}=40$

$\Rightarrow \ 46=40+40\alpha$

$\Rightarrow \ \alpha-\dfrac {6}{40}=\dfrac {3}{20}$

$\therefore \ \% \ NO_2=\dfrac {2\alpha}{1+\alpha}\times 100$

$=\dfrac {10}{23}\times 100+\dfrac {6}{23}\times 100 \times 26.08\%$

Hence, the correct option is

$\text{C}$

An unknown chlorohydrocarbon as

$3.55$ % of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, chlorine atoms present in

$1g$ of chlorohydrocarbon are: (Atomic wt. of

$Cl=35.5u$ ; Avogadro constant

$=6.023\times {10}^{23}{mol}^{-1}$ )

0%
$6.032\times {10}^{9}$
0%
$6.032\times {10}^{23}$
0%
$6.032\times {10}^{21}$
0%
$6.032\times {10}^{20}$

Explanation

An unknown chlorohydrocarbon has

3.55% of chlorine.
100 g of chlorohydrocarbon has

3.55 g of chlorine.
1 g of chlorohydrocarbon will have

$3.55\times \dfrac { 1 }{ 100 } =0.0355g$
Atomic wt. of

$Cl$ = 35.5g/mol
Number of moles of

$Cl=\dfrac { 0.0355g }{ 3.55g/mol } =0.001mol$
Number of atoms of

$Cl$ =

$0.001mol\times 6.023\times 10^{ 23 }mol^{ -1 }=6.023\times 10^{ 20 }$

$400mg$ capsule contains

$100mg$ of ferrous fumarate. The percentage of iron present in the capsule approximately is: (Mol mass of

$Fe({C}_{4}{H}_{2}{O}_{4})=170$ )

0%
$8.2$ %
0%
$25$ %
0%
$16$ %
0%
$50$ %

Explanation

Molar mass of ferrous fumarate

$(C_4H_2FeO_4)=169.9$

$g$

Atomic mass of Iron

$(Fe)=55.8$

$\therefore 169.9$

$g$ of

$C_4H_2FeO_4\rightarrow 55.8$

$g$ of

$Fe$

$\therefore 100$

$mg$ of

$C_4H_2FeO_4$ contain:

$=\cfrac{55.8 g}{169.9 g}\times 100$

$mg$

$=32.8$

$mg$

Amount of

$C_4H_2FeO_4$ in

$400$

$mg$ capsule

$=100$

$mg$

$\therefore$ Amount of iron in

$400$

$mg$ capsule

$=32.8$

$mg$

Percentage of iron in

$400$

$mg$ capusle

$=100\times \cfrac {Amount\quad of\quad Iron\quad in\quad capsule}{Total\quad mass\quad of\quad capsule}$

$=100\times \cfrac{32.8 mg}{400 mg}$

$=8.2\%$

Phosphoric acid is widely used in carbonated beverages, detergents, toothpaste and fertilizers. The mass percentages of

$H,\ P$ and

$O$ in phosphoric acid are

$H - 3.06\, P - 31.63\%, O - 65.31\%$ . The atomic masses

$H = 1, P = 31$ and

$O= 16$ .

0%
True
0%
False

Explanation

Molecular formula of phosphoric acid is

${ H }_{ 3 }{ PO }_{ 4 }$ .

$\therefore$ Molecular mass of

${ H }_{ 3 }{ PO }_{ 4 }=1\times 3+31\times 1+16\times 4=98$

$\therefore$ Mass percentage of

$H=\dfrac { 1\times 3 }{ 98 } \times 100$ %

$=3.06$ %

$\therefore$ Mass percentage of

$P=\dfrac { 31\times 1 }{ 98 } \times 100$ %

$=31.63$ %

$\therefore$ Mass percentage of

$O=\dfrac { 16\times 4 }{ 98 } \times 100$ %

$=65.31$ %

Answer is true.

$1g$ of a sample of brass om reacting with excess HCI produces

$120 ml$ of

${ H }_{ 2 }$ gas at STP. The percentage Zn in this sample of brass is: (At.wt.of Zn-=

$65.5g$ )

0%
$32\%$
0%
$35\%$
0%
$38\%$
0%
$40\%$

Explanation

$Zn+2HCl\to ZnCl_2+H_2$

Moles of

$H_2$ released

$=\dfrac {120}{22400}=0.0053$

$\therefore \$ Moles of

$Zn=\dfrac {0.0053}{20}=5.3\ mole$

$\therefore \ \%$ of

$Zn=\dfrac {5.3\times 10^{-3}\times 65.5}{1}\times 100$

$=35.7\ \%$

$500$ grams of water dissolved 2 moles of potassium sulphate the mass precentage of salt in soluton:

0%
$41\%$
0%
$60\%$
0%
$35\%$
0%
$48\%$

Explanation

Given that ,

$2$ moles of potassium sulphate is dissolved in

$500$ gm of water

we know that ,

Molar mass of potassium sulphate

$= 174g$

$\therefore$ Mass of potassium sulphate dissolved

$=2\times 174= 348\ g$

Mass percentage of salt

$= \dfrac{\text{Mass of solute }}{\text{Mass of solution}}\times 100 \text %= \dfrac{348}{500+348}\times 100 \text %=\dfrac{348}{848}\times 100 \text %=41.03\text %=41\text %$

The mass percentage composition of the elements in nitric acid are

$H = 1.59\%,$

$N = 22.22\%,$

$O = 76.19\%$ respectively.

$(H = 1, N = 14, O =16)$

0%
True
0%
False

Explanation

Molecular formula of nitric acid

$={ HNO }_{ 3 }$ .

$\therefore$ Molecular mass of

${ HNO }_{ 3 }=1\times 1+14\times 1+16\times 3=63$

$\therefore$ Mass percentage of

$H=\dfrac { 1\times 1 }{ 63 } \times 100$ %

$=1.59$ %

Mass percentage of

$N=\dfrac { 14\times 1 }{ 63 } \times 100$ %

$=1.59$ %

Mass percentage of

$O=\dfrac { 16\times 3 }{ 63 } \times 100$ %

$=76.19$ %

Answer is true.

Calculate the mass of iron which will be converted into its oxide

$(Fe_3O_4)$ by the action of

$8g$ of steam on it. [At mass of

$Fe = 56$ ]

$Fe + H_2O \rightarrow Fe_3O_4 + 4H_2$

0%
$21$ g
0%
$12$ g
0%
$42$ g
0%
$84$ g

Explanation

$3Fe+4H_{ 2 }O\rightarrow Fe_{ 3 }O_{ 4 }+4H_{ 2 }$

As per stiochiometry, Moles of

$H_2O$ =Moles of fe

$4\times \cfrac{8}{18} =\cfrac{m}{56}\times 3$

$m=21 g$

Therefore, mass of iron converted to

$Fe_3O_4$ is

$21g.$

What is the mass of the precipitate formed when

$50$ mL of

$16.9$ %

$(w/v)$ solution of

$AgN{ O }_{ 3 }$ is mixed with

$50$ mL of

$5.8$ %

$(w/v)$

$NaCl$ solution?

$(Ag = 107.8, N =14, O = 16, Na = 23, Cl= 35.5)$

0%
$7$
0%
$14$
0%
$28$
0%
$3.5$

Explanation

$AgNO_{ 3 }+NaCl\rightarrow AgCl\downarrow +NaNO_{ 3 }$

Moles of

$AgNO_{ 3 }=\cfrac{\cfrac{16.9\times 50}{100}}{170}=0.05$ moles

Moles of

$NaCl =\cfrac{\cfrac{5.8\times 50}{100}}{58.5} =0.05$ moles

Therefore, the moles of precipitate

$(AgCl)= 0.05$ moles

Mass of precipitate formed

$(AgCl) =0.05\times 143.5=7.165g$

A solution of phosphoric acid was made by dissolving 10.0 g

$H_3PO_4$

in 100.0 mL water. The resulting volume was 104 mL. What is the , mole fraction of the solution

0%
$1$
0%
$2$
0%
$0.01$
0%
$3$
0%
$1.5$

$0^{\circ}C$ the density of gaseous oxide at 2 bar is same as that of nitrogen at 5 bar and

$0^{\circ}C$ . The molar mass of the oxide is (assuming ideal behaviour):

0%
$70 g/mol$
0%
$140 g/mol$
0%
$28 g/mol$
0%
$60 g/mol$

Explanation

We know Density

$\rho= \cfrac {M_\rho}{RT} \longrightarrow (1)$

For the given data if

$M$ is the nolar mass of the gaseous oxide we have,

$\rho= \cfrac {M}{RT}.2$ or

$2= \cfrac {\rho RT}{M}\longrightarrow (2)$

Also for nitrogen

$5=\cfrac {\rho RT}{28}\longrightarrow (3)$

$(2)$ and

$(3) \Rightarrow \cfrac {5}{2}= \cfrac {M}{28}$

$M= \cfrac {5\times 28}{2}=70g/mol$

Calculate the mass percent of carbon in

${ C }_{ 2 }H_{ 5 }OH$ (Molar mass of ethanol=46.068g)

0%
$13.13\%$
0%
$88.79\%$
0%
$52.14\%$
0%
$34.73\%$

Explanation

Given : Mass of

$C_2H_5OH=46.08g$

Molecular mass

$=46.08g$

Percentage of carbon

$=\cfrac{\text{Mass of carbon}}{\text{Molecular mass}}\times 100\\=\cfrac{24}{46.08}\times 100\\=52.14\%$

Four one litre flasks are separately filled with gases

$O_2, F_2, CH_4$ and

$CO_2$ under same conditions.

The ratio of the number of molecules in these gases are:

0%
$2 : 2 : 4 : 3$
0%
$1 : 1 : 1 : 1$
0%
$1 : 2 : 3 : 4$
0%
$2 : 2 : 3 : 4$

Explanation

Avogadro's law: It states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

$\\ n_{1} : n_{2} : n_{3} : n_{4} = \cfrac{PV}{RT} : \cfrac{PV}{RT} : \cfrac{PV}{RT} : \cfrac{PV}{RT} = 1 : 1 : 1 : 1$

So, the ratio of no. of molecules

$= 1: 1: 1: 1$

The correct option is

$B.$

According to Avogadro's law the volume of a gas will ____ as _____ if ____ are held constant.

0%
increases, number of moles; P & T
0%
decreases, number of moles; P & T
0%
increases; T & P; number of moles
0%
decreases; P & T; number of moles

Explanation

According to Avogadro's law: Equal volume of all gases at same temperature and pressure will have same no. of molecules.

For a given mass of ideal gas,

Volume

$\propto$ Number of moles of the gas (if temperature and pressure are constant)

So, volume of the gas will increase as the number of moles if pressure (P) and temperature (T) are held constant.

A 1.50 g sample of an ore containing silver was dissolved,and all the

$Ag^{+}$ was converted to 0.125 g

$Ag_{2}S$ . What was the percentage of silver in the ore?

0%
14.23%
0%
8.27%
0%
10.8%
0%
7.2%

Explanation

Number of moles of

$Ag_{2}S=\cfrac{0.125 g}{\text{Molecular weight of } Ag_{2}S}=\cfrac{0.125}{247.8}=5 \times 10^{-4} mole$

Mass of silver

$= 5 \times 10^{4} mole \times 2 \times 108=0.108 g$

$\therefore$ Percentage of silver in ore

$=\cfrac{0.108}{1.5} \times 100 = 7.2 \%$

To convert molality into which of the following unit of concentration require density of the solution?

0%
Percentage weight by weight
0%
Percentage by volume
0%
Mole fraction
0%
Given all

The percentage by weight of:
a] C in carbon dioxide
b] Na in sodium carbonate
c] AI in aluminium nitride.

are

$27\%$ ,

$43.4\%$ ,

$65.85\%$ respectively
[C=12, O=16, H=1,Na=23, AI=27, N=14]

0%
True
0%
False

Explanation

Solution:- (A) True

As we know that,

$\text{percentage by weight} = \cfrac{\text{weight}}{\text{total weight}} \times 100$

Therefore,

(a)percentage by weight of

$C$ in carbon dioxide-

Weight of carbon in

$C{O}_{2} = 12 \; g$

Weight of

$C{O}_{2} = 44 \; g$

$\therefore$ percentage by weight

$= \cfrac{12}{44} \times 100 = 27.27 \% \approx 27 \%$

(b)percentage by weight of

$Na$ in sodium carbonate-

Weight of sodium

$= 23 \; g$

Weight of sodium in

${Na}_{2}C{O}_{3} = 2 \times 23 = 46 \; g$

Weight of

${Na}_{2}C{O}_{3} = 106 \; g$

$\therefore$ percentage by weight

$= \cfrac{46}{106} \times 100 = 43.4 \%$

(c)percentage by weight of

$Al$ in aluminium nitride-

Weight of aluminium in

$AlN = 27 \; g$

Weight of

$AlN = 41 \; g$

$\therefore$ percentage by weight

$= \cfrac{27}{41} \times 100 = 65.85 \%$

Hence the given statement is true.

A gaseous mixture contains

$CH_4$ and

$C_2H_6$ in equimolecular proportion. The weight of 2.27 litres of this mixture at STP is:

0%
3 g
0%
4.6 g
0%
1.6 g
0%
2.3 g

Explanation

Mol wt of

$CH_4=16\;g/mol$

Mol wt of

$C_2H_6=30\;g/mol$

Total

$=46\;g/mol$

$\Rightarrow 1$ Mol wt of

$CH_4=22.4L$

$1$ Mol wt of

$C_2H_6=22.4L$

$\Rightarrow 46\;g/mol$ of

$CH_4+C_2H_6=22.4+22.4$

$=44.8L$

$\Rightarrow 46\;g/mol=44.8L$

$\Rightarrow x=2.27L$

$\Rightarrow x=\dfrac{46\times 2.27}{44.8}=2.33g$

Hence, the answer is

$2.33g.$

Hence, the correct option is

$D$

25.5g of

$H_{2}O_{2}$ solution on decomposition gave 1.68L of

$O_{2}$ at STP. The percentage strength by weight of the solution is:

0%
$30$
0%
$10$
0%
$20$
0%
$25$

Explanation

Balanced equation for decomposition of

$H_2O_2$

$2H_2O_2\rightarrow 2H_2O + O_2$

2mol

$H_2O_2$ will produce 1 mol

$O_2$

At STP 1 mol

$O_2$ has volume = 22.4L

At STP 1.68L of

$O_2$ = 1.68/22.4 = 0.075 mol

$O_2$

Therefore the solution contained

$0.075\times2 = 0.15$ mol

$H_2O_2$

Molar mass

$H_2O_2 = 34g/mol$

Mass of

$H_2O_2$ decomposed

$= 0.15\times34 = 5.1$ g

$H_2O_2$ in 25.5g solution

100g solution contains :

$100/25.5\times5.1 = 20%$

The original solution contains

$20% m/m H_2O_2$

Hence, the correct option is

$C$

$100$ gm of an aq. solution of sugar contains

$40\%$ sugar by mass. How much water should be evaporated get

$50\%$ sugar solution by mass?

0%
$10g$
0%
$20g$
0%
$0.0g$
0%
$40g$

Explanation

$\underline{\text{Initial condition}}:-$

$40\%$ sugar (by mass),

Therefore, In

$100gm$ of an aqueous solution,

$40g$ sugar and

$60g$ water is present.

$\underline{\text{Final condition}}:-$

$50\%$ sugar (by mass),

Therefore, In

$100gm$ of an aqueous solution,

$50g$ sugar and

$50g$ water is present.

Amount of water to be evaporated

$=60-50=10\text{ }grams$

According to percentage weight arrange the following in descending order in the earth crust ?

0%
$O_2, Ca, Mg, S$
0%
$O_2, S, Mg, Ca$
0%
$S, Ca, Mg, O_2$
0%
$Ca, O_2, Mg, S$

Explanation

Oxygen is present in maximum concentration in earth's crust while

$s$ has less abundance,

So, abundance of availability of elements

$O_2>Ca>Mg>S.$

One mole sample of

$FeO$ is heated in air until it is completely converted into

$Fe_2O_3$ . The percentage increase in weight of the sample is?

[Given: Atomic mass of Fe = 56]

0%
$10.58\%$
0%
$20.28\%$
0%
$35.35\%$
0%
$11.11\%$

Explanation

Mass of

$1$ mol of

$FeO=(56+16)=72g$

$FeO+\cfrac{3}{4}O_2\longrightarrow \cfrac{1}{2}Fe_2O_3$

$1$ mol of

$FeO$ gives

$\cfrac{1}{2}$ mol

$Fe_2O_3$ .

$\therefore$ Mass of sample after conversion

$=\cfrac{1}{2}\times M_{Fe_2O_3}=\cfrac{1}{2}\times 160=80g$

$\therefore \%$ increase in weight

$=\cfrac{80-72}{72}\times 100=11.11\%$

Hence, the correct option is

$\text{D}$

Aqueous urea solution is 20% by mass of solution. Calculate percentage by mass of solvent.

0%
75%
0%
15%
0%
25%
0%
65%

Explanation

Let mass of the solution=

$100g$

Given

$20$ % urea=

$20g$ urea and

$80g$ of water

$\therefore$ Mass of urea by solvent

$=\cfrac {20}{80}\times 100$

$=25$ %

The percentage weight by weight of

$0.5$ M

$CCl_4$ solution in benzene is?
(Density of solution =

$1.4$ g/ml)

0%
$5.5\%$
0%
$8.25\%$
0%
$13\%$
0%
$20.06\%$

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 9 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 9 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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